If a parcel of air has a temperature of -1. 1c, what is saturation mixing ratio for this parcel?

Answer:

four

Explanation:

i said so

The sample of gas that is unreactive to flame tests and chemically reactive with limewater and bromothymol blue is Carbon dioxide.

Flame tests are performed to identify the unknown metals or metalloids.  Upon burning on the flame, a compound gives characteristic colour, that helps in identification.

Carbon dioxide extinguishes the flame itself when placed upon it. And it is known to turn the lime water milky when reacts. Carbon dioxide also changes the colour of bromothymol blue. The change in colour is observed due to certain chemical changes like change in pH, formation of carbonic acid and conversion into more acidic form.

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The value of ''n" in the hydrate formula is 10

Mass of anhydrous NA2B4O7 = 2.64g

Molar mass of NA2B4O7 = 201.224 g/mol

Number of Moles of anhydrous NA2B4O7 = Mass of NA2B4O7 / Molar mass of NA2B4O7

                                                     = 2.64 / 201.224

                                                     = 0.0131 mol

Mass of water in 5.00g of NA2B4O7 = 5.00 - 2.64

                                                              = 2.36 g

Moles of water = 2.36 / 18.015

                          = 0.1307 mol

The whole number ratio = moles of water / moles of anhydrous sodium tetra borate

                                           = 0.1307 / 0.0131

The whole number ratio = 10

Hence the value of n in the hydrate formula is 10.

NA2B4O7. 10H2O

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Answer:

25 gm would be left

Explanation:

45 hours is 3 half lives

200 gm  * 1/2 * 1/2 * 1/2 = 25 gm

According to the concept of half life , as 45 hours are equal to three half- lives after 3 half-lives 25 g of sodium-24 remains.

Half -life of a substance is defined as the time which is required for half of the quantity of a radioactive substance to get decayed.It is a term which is used in nuclear chemistry for describing how quickly unstable atoms undergo radioactive decay into other nuclear species by emitting particles or the time which is required for number of disintegrations per second of radioactive material to decrease by one half of its initial value.

In the given problem time is=45 hours and as 1 half  life=15 hours , for 45 hours=45/15=3 half lives .

Amount of sodium-24 after 3 half lives= 200×1/2×1/2×1/2=25 grams.

Therefore, after 45 hours 25 g of sodium-24 remains.

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Considering the definition of molar volume and density,the density of copper at 25°C is 8.96 [tex]\frac{g}{cm^{3} }[/tex].

Molar volume is an intensive property that indicates how much space one mole of a certain substance or compound occupies. In other words, the molar volume is the space occupied by one mole of a substance.

The expression to calculate the molar volume is:

Vm= V÷ n

where:

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density= mass÷ volume

The relationship between molar volume and density is reflected in the following expression, where the molar volume is calculated by dividing the molar mass (M) by the mass density (ρ):

Vm= M÷ρ

In this case, you know:

Replacing in the relationship between molar volume and density:

7.09 [tex]\frac{cm^{3} }{mol}[/tex]= 63.5 [tex]\frac{g}{mol}[/tex]÷ρ

Solving:

7.09 [tex]\frac{cm^{3} }{mol}[/tex]×ρ= 63.5 [tex]\frac{g}{mol}[/tex]

ρ= 63.5 [tex]\frac{g}{mol}[/tex]÷7.09 [tex]\frac{cm^{3} }{mol}[/tex]

ρ= 8.96 [tex]\frac{g}{cm^{3} }[/tex]

Finally, the density of copper at 25°C is 8.96 [tex]\frac{g}{cm^{3} }[/tex].

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pH value of a 0. 011 m naf solution is 8.57.

Solution

This problem uses the relationship between Kb and the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:

Kb = KaKb

1.00 x 10^-14 = 7.2 x 10^-4(x)

x = 1.39 x 10^-11

We now need to calculate the [OH¯] using the Kb expression:

1.39 x 10^-11 = x^2 / (0.30 - x)

The denominator can be neglected. Thus, x is 3.73 x 10^-6.

pOH = -log 3.73 x 10^-6 = 5.43

pH = 14-5.43 = 8.57

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The energy required to melt 20 g of water (ice) starting at 0° C is 6680 J.

Heat is a form of energy that is possessed by a system by virtue of its temperature. Its unit is Joule.

Enthalpy is the amount of the heat content of a system.

It is the amount of heat required to increase the temperature of a system by 1° C. It is an extensive property. Its unit is J/°C

Heat Capacity is given by

[tex]c= \frac{q}{\Delta T}[/tex]

where q = heat required

[tex]\Delta T[/tex] = change in temperature

Enthalpy of Fusion is the enthalpy change when 1 mole of a solid substance changes into its liquid state at its melting point.

Here, water in solid form (ice) melts into liquid form at 0°C

[tex]H_{2} O (s) \rightarrow H_{2} O(l) \\\Delta H_{fus } =334Jg^-^1[/tex]

We know,

[tex]\Delta H = \frac{m}{q}\\[/tex]

where m =  given mass

q = heat required

[tex]q = \Delta H \times m[/tex]

q = 334 [tex]Jg^-^1[/tex] x 20 g

    = 6680 J

Thus, the heat required to melt 20 g of water at 0°C is 6680 J

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The potential of a standard hydrogen electrode (s. h. e. ) be under the given conditions is - 0.029 V

Calculation ,

Formula used :

[tex]E_{cell}[/tex] = [tex]E_{cell}[/tex] ° - 0.059/n ㏒[tex]H_{2}[/tex]/ [tex][H^{+}]^{2}[/tex]

As we know that ,

[[tex]H^{+}[/tex]] = 0. 84 M

[tex]P_{H_{2}[/tex] = 2. 2 atm

Temperature ( T ) = 298  K

From reaction at electrode number of electrons involve ( n ) = 1

Standard electrode potential of  standard hydrogen electrode [tex]E_{cell}[/tex] ° = 0

Putting the value of all data in equation ( i ) , we get

[tex]E_{cell}[/tex] = 0  - 0.059/1 ㏒2. 2 / [tex](0. 84)^{2}[/tex] =  - 0.059 ㏒3.14 =  - 0.059×0.497

[tex]E_{cell}[/tex] = - 0.029 V

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At STP, the volume of water that is produced is 134.4 L.

Now we know that the equation of the reaction is;

2C6H6(g) + 15O2(g) = 12CO2(g) + 6H2O(g)

We can use the stoichiometry of the reaction to determine the volume of the water produced as follows;

2 mole of C6H6 produced 6 moles of water. If 1 mole of water occupies 22.4 L then 6 moles of water will occupy 6 * 22.4 L = 134.4 L

Thus, we can see that at STP, the volume of water that is produced is 134.4 L.

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Explanation:

harmful effects of ozone layer depletion on human being are as follows :

i. It increases in certain type of cancer.

ii. It increases in eyes cataracts.

iii. It increases in immune deficiency disorder.

iv. It changes the climate.

Answer:

single displacement

Explanation:

the metal replaces the hydrogen of the acid to for a salt and hydrogen gas

Answer: a. 65.37 mL
b. 5.3 cm
c. 2.08 km (plus) or 1.64 km (minus)
d. 18°C (plus) or 16°C (minus)

Explanation:

The way significant figures work when you're adding or subtracting, is you go to the answer with the least amount of decimal places. Take a. for example; you have 65.316 (3 places after the decimal) and 0.05 (2 places after the decimal place). Based off of this, you would round your answer to two places after the decimal. And when you have a whole number, like in d. you round to the nearest whole number, since there aren't any decimal places to round your answer to.

Answer:

synergy

Explanation:

when two substances combine effects to be greater than just their sum of their individual effects

When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.

The precipitate is the solid concentration of a substance that is collected over a solution.

First, we determine the concentration of magnesium and hydroxide

(Mg2+) = 7.00 × 10⁻⁴

(OH−) = 5.00 × 10⁻⁵

Now, we calculate the solubility quotient

Qc = (Mg2+) (OH−) ²

Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²

Qc = 1.75 x 10⁻¹²

The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.

Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.

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Taking into account the reaction stoichiometry and the definition of STP, 72.8 L of CO₂ gas is produced once the gas cools to STP.

In first place, the balanced reaction is:

CaCO₃  → CaO + CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CaCO₃ form 1 mole of CO₂, 3.25 moles of CaCO₃ form how many moles of CO₂?

[tex]moles of CO_{2} =\frac{3.25 moles of CaCO_{3}x 1 mole of CO_{2}}{1 moleof CaCO_{3}}[/tex]

moles of CO₂= 3.25 moles

Then, 3.25 moles of CO₂ are formed when 3.25 moles of CaCO₃ decomposes.

Now, you can apply the following rule of three: if by definition of STP conditions 1 mole of CO₂ occupies a volume of 22.4 liters, 3.25 moles occupies how much volume?

[tex]volume= \frac{3.25 molesx22.4 L}{1 mole}[/tex]

volume= 72.8 L

Finally, 72.8 L of CO₂ gas is produced once the gas cools to STP.

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Succinate dehydrogenase is the citric acid cycle considered part of aerobic metabolism even though oxygen does not appear.

The respiratory complex II, also known as succinate dehydrogenase (SDH), succinate-coenzyme Q reductase (SQR), or SDH, is an enzyme complex that is present in the inner mitochondrial membrane of eukaryotic and numerous bacterial cells. SDH converts succinate to fumarate as part of the citric acid cycle. SDH shares structural similarities with fumarate reductase, an enzyme that catalyzes the reverse process during anaerobic respiration in bacteria. (1997, Hagerhall). Succinic semialdehyde dehydrogenase impairment is brought on by ALDH5A1 gene mutations. Instructions for creating the succinic semialdehyde dehydrogenase enzyme are found in the ALDH5A1 gene.

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A monosaccharide that contains 6 carbon atoms, one of which is in an aldehyde group, is classified an Aldohexose.

Monosaccharides of specific sizes may be indicated by names composed of a stem denoting the number of carbon atoms and the suffix -ose.

For example, the terms triose, pentose, and hexose signify monosaccharides with, respectively, three, four, five, and six carbon atoms. Monosaccharides are also classified as aldoses or ketoses.

Those monosaccharides that contain an aldehyde functional group are called aldoses; those containing a ketone functional group on the second carbon atom are ketoses. Combining these classification systems gives general names that indicate both the type of carbonyl group and the number of carbon atoms in a molecule.

. Glucose and fructose are specific examples of an aldohexose and a ketohexose, respectively.

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Student e accounted for the equivalent weight found for succinic acid by analyzing its titration with [tex]NaOH[/tex] (aq) and concluding that it is diprotic.

One mole of succinic acid required two mole of base to neutralize completely. In this titration phenolphthalein indicator is used to observe the completion of the reaction. In an aqueous solution, succinic acid readily ionize to form succenate. Succinic acid undergoes two successive deprotonation reaction.

The succinic acid results in succinate ( dicarboxylic acid dianion ) which is form by removal of a proton from both carboxy groups of succinic acid.

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Nuclear decay is a radioactive decay where energy is released by unstable nuclei. After thorium undergoes decay, actinium-228 will be the final product. Thus, option a is correct.

Alpha radioactive decay is the loss of two protons and neutrons from the element that results in a decrease of the atomic mass by 4 and atomic number of the element by two.

²³² Th₉₀ → ²²⁸Ra₈₈ + ⁴α₂

Beta radioactive decay is the increase of the atomic number but not the atomic mass of the element.

²²⁸Ra₈₈ → ²²⁸Ac⁸⁹ + β⁻

Therefore, option a. actinium-228 is the final product of the decay.

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Your question is incomplete, but most probably your full question was, In the initial sequence of thorium-232 decay, an alpha particle is emitted followed by a beta particle. What is the product of these two decay steps?

²³² Th → α particle + β particle + ?

Answer: 5kg

Explanation: formula K=1/2mv^2

50=10m

Answer:

C

Explanation:

KE = 1/2 m v^2

KE *2 / v^2   = m

50 (2)/ 10^2   = m =  1 kg

The two ways were,

A cathode-ray tube, often known as a CRT, is a vacuum tube that houses one or more electron cannons that produce electron beams that are then bent to produce pictures on a phosphorescent screen. The visuals might be pictures (from a television or computer display), electrical waveforms (from an oscilloscope), radar targets, or other occurrences.

Tomson came to the conclusion that atoms can be divided into their constituent corpuscles (atoms are made up of smaller particles).

The electron was discovered in 1897 by J. J. Thomson.

His "plum pudding" concept from 1904 postulated that the positive charge contains embedded electrons.

With this model, he disproved his first theory (the atom was composed of immaterial vortices).

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Answer:

A. 2.74g is the correct answer

Explanation:

[tex]average \: mass = \frac{(mass \: pre - 1982 \times abundance) + (mass \: post - 1982 \times abundance}{100 } \\ ave = \frac{ (3.1g \times 40.0) + (2.5g \times 60.0)}{100} \\ ave = \frac{124 + 150}{100} \\ ave = \frac{274}{10} = 2.74g[/tex]

where, the sum of abundance always have to be 100%

Thus, 40.0 +60.0= 100.0

In the polymerization reaction, the lone pair electrons on the NH₂ groups of hexanediamine attack the C=O groups of the dicarboxylic acid in a nucleophilic substitution reaction as shown in the image.

Hydroxide is added to remove any H⁺ ions present and keep the hexanediamine in the deprotonated form, so that the NH₂ lone pair electrons are available for reaction.

If hydroxide is not added, the NH₂ groups will get protonated by H⁺  ions present to give NH₃⁺ groups, which cannot react.

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The concentration of [tex]NaCl[/tex] solute particles in water will provide an isotonic eyedrop solution is 0.9% w/v

The isotonic solution is calculated by multiply the quantity of each drug in the prescription by its sodium chloride equivalent E , and subtract this value from the concentration of sodium chloride which is isotonic with body fluids . For a solution to be termed isotonic ( equal tone ) it must have the same osmotic pressure as a specific bodily fluid.

Example : The concentration of [tex]NaCl[/tex] solute particles in water will provide an isotonic eyedrop solution is 0.9% w/v

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If we takes 11. 2 kj of energy to raise the temperature of 145 g of benzene from 23. 0°c to 68. 0°c, then the specific heat of benzene is

1716.475 J/kg.°C.

Specific Heat capacity is given as

Q = cm(t₂-t₁) ................[tex]equation 1[/tex]

Where Q = quantity of heat, c = specific heat of benzene, m = mass of benzene, t₁  = initial temperature, t₂ = final temperature.

[tex]C[/tex] = [tex]\frac{Q}{m_{(t2-t1)} } ...........equation 2[/tex]

Given:

[tex]Q = 11.2 kJ = 11200 J \\\\m= 145\\\\g = 0.145 kg\\\\t_{2} = 68.0 °C\\\\t_{1} = 23.0 °C[/tex]

Substitute into equation 2.

[tex]C = \frac{11200}{[0.145(68-23)]}\\\\ C= \frac{11200}{6.525}\\\\ C= 1716.475 J/kg °C[/tex]

Hence the specific heat of benzene = [tex]1716.475 J/kg °C[/tex]

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option (c) emulsion is the right answer

Emulsion is a classification of milk.

In physical chemistry, an emulsion is a combination of two or more liquids in which one of the liquids is present as microscopic or ultramicroscopic droplets dispersed throughout the other.

Emulsions form as a result of the cleansing action of soaps.

(ii) The emulsification process is how lipids are broken down in the intestines.

(iii) Disinfectants and antiseptics combine with water to generate emulsions.

(iv) The emulsification technique is utilized to create medications.

When two immiscible liquids, such as oil and water, are stirred together with an emulsifier—which might be a protein, phospholipid, or even a nanoparticle—emulsion is created.

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The work done on the surroundings by the expansion of a gas is w = –pdv .The statement is true.

The work is the energy required to move a system or body or anything against a force is called work. The word done = force×distance

Work done in an isobaric process is given by

W = PΔV

Where W = Work done

P =pressure

ΔV = change in volume

In a free expansion , the gas is allowed to expand into a vacuum.

This happens so quickly , so there is no heat transferred . No work is done , because the gas does not display anything.

W = PΔV

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The molar solubility is 7.4×[tex]10^{-3}[/tex] M and the solubility is  7.4×[tex]10^{-3}[/tex] g/L .

Calculation ,

The dissociation of silver bromide is given as ,

[tex]AgBr[/tex] → [tex]Ag ^{+}[/tex] + [tex]Br^{-}[/tex]

S  

 -          S        S        

Ksp =  [[tex]Ag ^{+}[/tex] ] [ [tex]Br^{-}[/tex] ]  =  [S] [ S ] = [tex]S^{2}[/tex]

S = √ Ksp = √ 5. 5×[tex]10^{-5}[/tex] = 7.4×[tex]10^{-3}[/tex]

The solubility =7.4×[tex]10^{-3}[/tex] g/L

The molar solubility is the solubility of one mole of the substance.

Since ,  one mole of [tex]AgBr[/tex] is dissociates and form one mole of each  [tex]Ag ^{+}[/tex] and [tex]Br^{-}[/tex] ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4×[tex]10^{-3}[/tex] mol/L = 7.4×[tex]10^{-3}[/tex] M

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